[QUOTE=rogue;268587]It is an unfortunate fact that Don is still trying to peddle his nonsense. He gets a couple of less competent mathematicians to agree with him and that bolsters his thinking regarding his proof, but if those mathematicians were to congregate with more competent people who see the flaws, then they would easily be convinced that their initial judgement was wrong.[/QUOTE]

Yes and no. Your argument assumes that the cranks are [b]willing[/b]
to accept critiques of their work and then to adjust their thinking accordingly.
Experience has shown that this is usually far from true. They are [b]willfully ignorant[/b]. They are blind to outside persuasion.

Pots and Kettles
 

[QUOTE=R.D. Silverman;268646]They are blind to outside persuasion.[/QUOTE]

Can't think of anyone around here who fits that description.
Maybe you can?

[QUOTE=R.D. Silverman;268646]Yes and no. Your argument assumes that the cranks are [b]willing[/b]
to accept critiques of their work and then to adjust their thinking accordingly.
Experience has shown that this is usually far from true. They are [b]willfully ignorant[/b]. They are blind to outside persuasion.[/QUOTE]

I know that nobody could convince him that he is correct, but if others can be convinced that he is wrong, then he will have fewer places to espouse his garbage.

[QUOTE=rogue;268652]I know that nobody could convince him that he is correct, but if others can be convinced that he is wrong, then he will have fewer places to espouse his garbage.[/QUOTE]To be perhaps more precise: if others can be convinced that he (Blazys) is wrong then, although he will have exactly the same number of places to espouse his garbage, he will be noticeably less likely successfully to delude the uninitiated.

Paul

[QUOTE=xilman;268679]To be perhaps more precise: if others can be convinced that he (Blazys) is wrong then, although he will have exactly the same number of places to espouse his garbage, he will be noticeably less likely successfully to delude the uninitiated.

Paul[/QUOTE]

The problem with this is that Don's psychological payoffs for peddling his stuff are higher than your psychological payoffs for preventing it by hunting down his threads and making pointers that his argument has nothing to do with mathematics.

Any other suggestions for changing the psychological equations here?

A letter from the frustrated author of a journal paper
 

Editor's Note: It seems appropriate, in this issue of JSS containing the fndings of our annual Top Scholars/Institutions study, to pay tribute to the persistent authors who make a journal like this, and a study like that, possible. In their honor, we dedicate the following humorous, anonymously-authored, letter!

[url]http://www.gruppoeratostene.com/articoli/Mersenne.pdf[/url]

seems to make quite a few mistakes:

[QUOTE]Euler showed:
Theorem: If k>1 and p=4k+3 is prime, then 2p+1 is prime if and only if 2p = 1 (mod 2p+1)[/QUOTE]

problem: 2p=-1 mod (2p+1) always so this says Euler stated that if p=4k+3 is prime 2p+1 is never prime. but likely a typo.


[QUOTE]Lemma 2
if p=4k+3 or p=4k-1 a prime number and 2p+1 is a prime number then the Mersenne number Mp=
2^p-1 is composite[/QUOTE]

problem: it's the one so far I've found where they give no demonstration. Also I notice that if lemma one was true half of the scenarios they talk about are already impossible.

[QUOTE]Lemma 3
Let be p = 4k-1 a prime number, with k>0 and k = 0 mod 3 then Mp=2^p-1 isn't a prime number .
Dem.
It follows from Lemma 2 [/QUOTE]

says the proof relies on on a previous lemma with no demo. It as well has no other supporting evidence given.

[QUOTE]Lemma 4 equivalent to Lemma 3
If p = 4k-1 is a prime number, if are true two conditions:
a. (p+1) mod 4 = 0
b. ((p+1)/4) mod 3 = 0
then Mp is composite
Examples:
p=11=4*3-1 k=3*1 Mp=2^11-1 not prime
p=23=4*6-1 k=3*2 Mp=2^23-1 not prime
p=47=4*12-1 k=3*4 Mp=2^47-1 not prime[/QUOTE]

equivalent to a lemma with no proof always nice , also some may comment on the low amount of examples for such a claim without an algebraic proof.

[QUOTE]
Lemma 5
If W =(4k+5)/3 is an odd with k>1 and W>3, if W is a Wagstaff's prime number then k = Mp is a
Mersenne's prime Number.
Dem.
If W is a Wagstaff's prime number then w=(2^p+1)/3 an p is a prime number.
Mp= 2^p-1 so [B]W=(Mp+2)/3[/B] but from Lemma 1 Mp=4k+3 then W=(4k+5)/3
If p is a prime number and W a Wagstaff prime number then k integer is a Mersenne's prime
number if 4k+5/3 is integer. [/QUOTE]

section highlighted is false on what they say if k=M[SUB]p[/SUB] then 4k+5 = M[SUB]p+2[/SUB]+2 not M[SUB]p[/SUB] +2.

[QUOTE]Theorem MMp
MMp is prime if Mp is prime and exists a k>=0 [B]such that (Mp*3-5)/4[/B] must be integer.
Dem.
Let be W=(2^p+1)/3 a Wagstaff's prime number, Mp=2^p-1 a Mersenne's prime number.
From Lemma 1 is Mp=4k+3
From Lemma 5 a Wagstaff's prime number is W=(Mp+2)/3=(4k+5)/3 or k=Mp=(3W-5)/4[/QUOTE]

anyone else figure out the missing part of the equation highlighted?

[QUOTE=science_man_88;273338]lemma one
[/QUOTE]

sorry theorem.

they talk (with a lot of typing mistakes and redundant things, for example p being 4k+3 is the same as p being 4k-1) about Sophie Germain lemma, "[B]IF p is prime and of the 4k+3 form[/B] (is 3 mod 4), [B]AND 2p+1 is prime, then 2p+1 divides 2^p-1[/B]". This is true in both directions, but we are interested only in the first part. If you need more examples, a simple pari routine like here will reveal all (let me find it, I have it somewhere):

[CODE]
elimSGp3m4(start,stop,file,pflag)=
{
p=start;
while(p<stop,
until(p%4==3 && isprime(2*p+1),
p=nextprime(p+1)
);
if(p<stop,
if(pflag, printf("%d%c",p,13));
write(file,p","2*p+1)
)
);
}
[/CODE]running it from 2 to 10^9 will produce something like:

[CODE]
11,23
23,47
83,167
131,263
179,359
191,383
239,479
251,503
359,719

... millions other lines

999973883,1999947767
999973979,1999947959
999974711,1999949423
999975299,1999950599
999976079,1999952159
999977399,1999954799
999977711,1999955423
999979439,1999958879
999980339,1999960679
999980603,1999961207
999980879,1999961759
999981863,1999963727
999982499,1999964999
999983723,1999967447
999983819,1999967639
999984899,1999969799
999984971,1999969943
999985391,1999970783
999987503,1999975007
999987743,1999975487
999989219,1999978439
999990851,1999981703
999992639,1999985279
999993083,1999986167
999993623,1999987247
999993719,1999987439
999995291,1999990583
999996131,1999992263
999996611,1999993223
999997571,1999995143
999998423,1999996847
999998903,1999997807
999999191,1999998383
[/CODE]The format is exponent, comma, factor of M(exponent).

I think the first thing GIMPS ever did was to create a list of prime exponents (candidates) and eliminate from this list the SG primes which are 3 (mod 4), about 1/3 of the exponents are gone by this lemma. The rest, well... TF, P-1, ECM, LL... :smile:

The rest of the article is gibberish.
[edit] and lemma 3 (equivalent lemma 4): if p=3 mod 4, then p+1 is 0 mod 4 always, so for (p+1)/4 to be 0 mod 3, then p+1 is 0 mod 12, that is, p is 11 mod 12. Lemma implies that all p of the form 11 mod 12 primes give composite mersenne. A bit fishy, don't you think? This fails epically for p=107.

Couldn't stop myself, went to the article.

[QUOTE]
So the authors say:
M3 prime MM3 prime!
M5 prime MM5 NOT prime
M7 prime MM7 prime!
M13 prime MM13 NOT prime
M17 prime MM17 NOT prime
M19 prime MM19 NOT prime
[/QUOTE]

well, for his general knowledge, mm5 is m31, which is prime...

Quoting LaurV,
Quote:
[QUOTE]
...certainly our don blazys totally lost his chance to prove it and take the money and the glory... :razz:

[/QUOTE]

Things like money and glory are meaningless to me.

However, if you Google search "Beal's Conjecture Proof",
then you will find that Google ranks my proof at the very top!

That's because it's both true and correct!

Doesn't that make you happy?

Don.